#  \sum_{n=1}^\infty (n^2 * x^n)/(2n)

This power series converges for -1<x<1 to the function f(x) = x/(2(1-x)^2).

Hello!

The given sum is a power series, and it has the form sum_( n = 1 )^(oo) a_n x^n  for a_n = n^2 / ( 2 n ) = n / 2 .

The question is for what values of x  the series converges and what is its sum. The ratio test for a power series shows that this series converges for -1 lt x lt 1 :  | a_( n + 1 ) / a_n | = ( n + 1 ) / n -gt 1 .  Therefore, the radius of convergence is 1. Note that for x = -1 and x = 1, the series diverges, because its term does not tend to zero.

Now is time to determine the value of this sum. Denote it as f ( x ) ;  then

f ( x ) / x = 1/2 sum_( n = 1 )^(oo) n x^( n - 1 ) .

It is known that we may integrate power series term-wise inside its interval of convergence, so we obtain

int ( f ( x ) ) / x dx = 1/2 sum_( n = 1 )^(oo) x^n = 1/2 * x / ( 1 - x ) = 1/2 ( -1 + 1/(1-x) ).

From this, we get ( f ( x ) ) / x = ( 1 / 2 ( -1 + 1/(1-x) ))' = 1/(2(1-x)^2) , and finally
f(x) = x/(2(1-x)^2) .